Question

Suppose that the electron had no spin and that the Pauli exclusion principle still held .Which, if any, of the present noble gases would remain in that category?

Short answer

The noble gas that would remain in that category is Argon.

Step-by-step solution

The given data

a) The electron has no spin.

b) The Pauli’s exclusion principle is still held.

Understanding the concept of the spin degree of freedom:

According to Pauli's Exclusion Principle, no two electrons in the same atom may have the same values for all four of their quantum numbers.

After the removal of the spin value, the spin degree of freedom becomes half of the electron states for each shell would become half as Pauli's Exclusion Principle is still applied. Now, comparing the electron states after the removal of other electron states from the shell and before removal, we can get the noble gas that maintains its configuration.

Calculation of the configuration of the noble gas that is maintained:

Without the spin degree of freedom, the number of available electron states for each shell would be reduced by half. So, the values of for the noble gas elements would become half of what they are now: $\mathrm{Z}=1,5,9,18,27\mathrm{and}43$.

Of this set of numbers, the only one that coincides with one of the familiar noble gas atomic numbers $\left(Z=2,10,18,36,54and86\right)\mathrm{is}18$.

Hence, the noble gas that would be the only one to remain “noble” is Argon.